{"id":26103,"date":"2019-03-15T19:01:38","date_gmt":"2019-03-15T13:31:38","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=26103"},"modified":"2019-09-11T15:53:35","modified_gmt":"2019-09-11T10:23:35","slug":"time-speed-questions-for-rrb-ntpc-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/time-speed-questions-for-rrb-ntpc-pdf\/","title":{"rendered":"Time & Speed Questions for RRB NTPC PDF"},"content":{"rendered":"

Time & Speed Questions for RRB NTPC PDF<\/span><\/h1>\n

Download RRB NTPC Time & Speed Questions and Answers PDF. Top 25 RRB NTPC\u00a0 Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam<\/p>\n

Download Time & Speed Questions for RRB NTPC<\/a><\/p>\n

RRB NTPC Previous Papers [Download PDF]<\/a><\/p>\n

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Question 1:\u00a0<\/b>P can do a work in 10 days, Q can do the same work in 20 days and after working for 2 days, C joins the work and after 2 days work has been completed. How many does C alone take to complete the work ?<\/p>\n

a)\u00a04<\/p>\n

b)\u00a03<\/p>\n

c)\u00a05<\/p>\n

d)\u00a06<\/p>\n

Question 2:\u00a0<\/b>A and B can do a piece of work individually in 14 hours and 21 hours respectively. A started the work and will be assisted by B on every alternate hour. Then find the time in which the total work will be completed?<\/p>\n

a)\u00a010 hours 40 minutes<\/p>\n

b)\u00a012 hours 20 minutes<\/p>\n

c)\u00a011 hours 24 minutes<\/p>\n

d)\u00a014 hours 40 minutes<\/p>\n

Question 3:\u00a0<\/b>A and B can do a piece of work individually in 12 hours and 20 hours respectively. They work on alternate hours starting with A. Then in how many hours, will the total work be completed?<\/p>\n

a)\u00a012 hr 48 min<\/p>\n

b)\u00a014 hr 48 min<\/p>\n

c)\u00a016 hr 36 min<\/p>\n

d)\u00a014 hr 36 min<\/p>\n

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Question 4:\u00a0<\/b>A,B and C can do a piece of work individually in 12,16 and 24 days respectively. They all started the work together but after 4 days, A and B left. Then, in how many days total work will be completed?<\/p>\n

a)\u00a04 days<\/p>\n

b)\u00a06 days<\/p>\n

c)\u00a010 days<\/p>\n

d)\u00a012 days<\/p>\n

Question 5:\u00a0<\/b>The speed of boat upstream and speed of boat in still water are in the ratio 3:4. If speed of stream is 24 km\/hr, then find the speed of boat downstream.<\/p>\n

a)\u00a0100 km\/hr<\/p>\n

b)\u00a0120 km\/hr<\/p>\n

c)\u00a0180 km\/hr<\/p>\n

d)\u00a0160 km\/hr<\/p>\n

Question 6:\u00a0<\/b>Two trains are running on two parallel tracks at 56 km\/hr and 38 km\/hr. The faster train passes a man standing in slower train in 24 seconds. Then find the length of the faster train.<\/p>\n

a)\u00a0100 m<\/p>\n

b)\u00a0160 m<\/p>\n

c)\u00a0180 m<\/p>\n

d)\u00a0120 m<\/p>\n

Question 7:\u00a0<\/b>A bus travels at 75 km\/hr without stoppages and if stops are included, the speed is reduced to 45 km\/hr. Then, how many minutes per hour does the bus stop?<\/p>\n

a)\u00a016 min<\/p>\n

b)\u00a022 min<\/p>\n

c)\u00a024 min<\/p>\n

d)\u00a032 min<\/p>\n

Question 8:\u00a0<\/b>The ratio of speed of boat upstream and speed of boat downstream is 5:14. If speed of stream is 18 km\/hr, then find the speed of boat in still water.<\/p>\n

a)\u00a034 km\/hr<\/p>\n

b)\u00a028 km\/hr<\/p>\n

c)\u00a038 km\/hr<\/p>\n

d)\u00a024 km\/hr<\/p>\n

Question 9:\u00a0<\/b>A train travels a certain distance at 60 km\/hr and returns at 30 km\/hr. Find the average speed.<\/p>\n

a)\u00a035 km\/hr<\/p>\n

b)\u00a040 km\/hr<\/p>\n

c)\u00a050 km\/hr<\/p>\n

d)\u00a045 km\/hr<\/p>\n

Question 10:\u00a0<\/b>A person travels from his home to his office at a speed of 40 km\/hr and returns at a speed of 60 km\/hr.<\/p>\n

a)\u00a042 km\/hr<\/p>\n

b)\u00a036 km\/hr<\/p>\n

c)\u00a024 km\/hr<\/p>\n

d)\u00a048 km\/hr<\/p>\n

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Question 11:\u00a0<\/b>A train travelled a distance of 480 km at uniform speed. If the speed is reduced by 8 km\/hr, it would travel 3 hours late. Then, find initial speed of the train.<\/p>\n

a)\u00a032 km\/hr<\/p>\n

b)\u00a048 km\/hr<\/p>\n

c)\u00a056 km\/hr<\/p>\n

d)\u00a040 km\/hr<\/p>\n

Question 12:\u00a0<\/b>A train travels a distance of 240 km at uniform speed. Due to breakdown, then speed is reduced by 10 km\/hr and hence it travels the destination 4 hours more. Find the initial speed of the train.<\/p>\n

a)\u00a025 km\/hr<\/p>\n

b)\u00a030 km\/hr<\/p>\n

c)\u00a035 km\/hr<\/p>\n

d)\u00a040 km\/hr<\/p>\n

Question 13:\u00a0<\/b>If speed of boat in still water is double of speed of boat upstream, then find the ratio of speed of boat downstream to that of current.<\/p>\n

a)\u00a02:3<\/p>\n

b)\u00a03:2<\/p>\n

c)\u00a04:3<\/p>\n

d)\u00a03:1<\/p>\n

Question 14:\u00a0<\/b>A boat can travel at 54 km\/hr downstream. If the speed of stream is 14 km\/hr, then find the speed of boat upstream.<\/p>\n

a)\u00a024 km\/hr<\/p>\n

b)\u00a036 km\/hr<\/p>\n

c)\u00a026 km\/hr<\/p>\n

d)\u00a038 km\/hr<\/p>\n

Question 15:\u00a0<\/b>If speed of boat in still water is triple of speed of current, then find the ratio between speed of boat upstream to that of downstream.<\/p>\n

a)\u00a01:2<\/p>\n

b)\u00a02:3<\/p>\n

c)\u00a03:4<\/p>\n

d)\u00a03:2<\/p>\n

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Question 16:\u00a0<\/b>If the ratio of speed of boat upstream to that of downstream is 1:2, then find the ratio of speed of boat in still water to speed of current.<\/p>\n

a)\u00a02:1<\/p>\n

b)\u00a03:1<\/p>\n

c)\u00a03:2<\/p>\n

d)\u00a04:3<\/p>\n

Question 17:\u00a0<\/b>A train crosses a platform of length 180 m in 16 seconds. It crosses a man standing on the platform in 4 seconds running at same speed. Then, find the length of the train.<\/p>\n

a)\u00a080 m<\/p>\n

b)\u00a060 m<\/p>\n

c)\u00a0120 m<\/p>\n

d)\u00a0100 m<\/p>\n

Question 18:\u00a0<\/b>A train of length 120 m crosses a platform of length 160 m in 14 seconds. Then, what is the speed of the train?<\/p>\n

a)\u00a072 km\/hr<\/p>\n

b)\u00a084 km\/hr<\/p>\n

c)\u00a020 km\/hr<\/p>\n

d)\u00a054 km\/hr<\/p>\n

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Question 19:\u00a0<\/b>Find the length of the train travelling with 90 kmph if it crossed a 600m bridge in 56 seconds ?<\/p>\n

a)\u00a0950 m<\/p>\n

b)\u00a0900 m<\/p>\n

c)\u00a0850 m<\/p>\n

d)\u00a0800 m<\/p>\n

Question 20:\u00a0<\/b>Find the speed of a train of length 600m, which crossed a man standing on the platform in 24 seconds ?<\/p>\n

a)\u00a040 kmph<\/p>\n

b)\u00a080 kmph<\/p>\n

c)\u00a045 kmph<\/p>\n

d)\u00a090 kmph<\/p>\n

Question 21:\u00a0<\/b>6 children can eat 48 chocolates in 12 days. Then how many children can eat 72 chocolates in 6 days ?<\/p>\n

a)\u00a014<\/p>\n

b)\u00a016<\/p>\n

c)\u00a018<\/p>\n

d)\u00a020<\/p>\n

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Question 22:\u00a0<\/b>Two trains X and Y are travelling in the same direction. Their lengths are 850m and 950m respectively. They took 6 mins to cross each other. Find the speed of train X if the speed of train Y is 90 kmph ?<\/p>\n

a)\u00a072 kmph<\/p>\n

b)\u00a074 kmph<\/p>\n

c)\u00a076 kmph<\/p>\n

d)\u00a078 kmph<\/p>\n

Question 23:\u00a0<\/b>Find the speed of the flight if it takes 6 hrs 30 mins to travel 2,210 km ?<\/p>\n

a)\u00a0280 kmph<\/p>\n

b)\u00a0300 kmph<\/p>\n

c)\u00a0320 kmph<\/p>\n

d)\u00a0340 kmph<\/p>\n

Question 24:\u00a0<\/b>A car travels x km at 60 kmph and then 2x km at 40 kmph. Find its average speed for entire distance ?<\/p>\n

a)\u00a045 kmph<\/p>\n

b)\u00a048 kmph<\/p>\n

c)\u00a056 kmph<\/p>\n

d)\u00a060 kmph<\/p>\n

Question 25:\u00a0<\/b>A train takes 50 seconds to cross a bridge of length 200m.Find the length of the train if the speed of the train is 50.4kmph ?<\/p>\n

a)\u00a0475<\/p>\n

b)\u00a0400<\/p>\n

c)\u00a0250<\/p>\n

d)\u00a0500<\/p>\n

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Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

LCM of 10 and 20 is 20 and so total units of work has to be done.
\nEach day P can do 2 units of work
\nEach day Q can do 1 unit of work
\nAfter 2 days 6 units of work is done so 14 units is left
\nGiven in next 2 days work has been completed along with C.
\nSo 2(1+2+x)=14
\n3+x=7
\nx=4
\nC does 4 units of work each day and so it takes 5 days for him to complete the whole work.<\/p>\n

2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let the total work be 42 units (LCM of 14 and 21)
\nEfficiency of A = 42\/14 = 3 units\/hour
\nEfficiency of B = 42\/21 = 2 units\/hour
\nA is assisted by B on every alternate hour
\nA can complete 3 units of work in first hour
\nA and B can complete 5 units of work in second hour.
\nHence, 8 units of work will be completed in 2 hours.
\nThen, 40 units of work will be completed in 10 hours.
\nRemaining work = 2 units.
\nA can do 2 units of work in 2\/3rd of an hour = $\\dfrac{2}{3}\\times60 = 40$ minutes
\nTherefore, Total work will be completed in 10 hours 40 minutes.<\/p>\n

3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the total work be 60 units (LCM of 12 and 20)
\nEfficiency of A = 60\/12 = 5 units\/hr
\nEfficiency of B = 60\/20 = 3 units\/hr
\nThey work on alternate days.
\nThen, A and B can do 8 units in 2 hours.
\n\u21d2 56 units of work will be completed in 14 hours.
\nRemaining work = 4 units
\nA can complete 5 units in 1 hour
\nThen, 4 units of work can be completed by A in 4\/5th of an hour = $\\dfrac{4}{5}\\times60 = 48 min
\nTherefore, Total work will be completed in 14 hours 48 min.<\/p>\n

4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let the total work be 48 units (LCM of 12,16 and 24).
\nEfficiency of A = 48\/12 = 4 units\/day
\nEfficiency of B = 48\/16 = 3 units\/day
\nEfficiency of C = 48\/24 = 2 units\/day
\nThen, Total work done by A,B,C in one day = 9 units
\nIn 4 days, 36 units of work will be completed.
\nRemaining work = 48-36 = 12 units.
\n12 units will be completed by C in 12\/2 = 6 days.
\nTherefore, Total work will be done in 4+6 = 10 days<\/p>\n

5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the speed of boat upstream be 3x km\/hr
\nSpeed of boat in still water = 4x km\/hr
\nThen speed of stream = 4x-3x = x km\/hr
\nGiven x = 24 km\/hr
\nThen, Speed of boat downstream = 4x+x = 5x km\/hr = 5*24 = 120 km\/hr<\/p>\n

6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Relative speed $= 56-38 = 18 km\/hr = 18\\times\\dfrac{5}{18} = 5 m\/sec$
\nFaster train crosses a man in slower train in 24 seconds.
\nLength of the faster train = 5*24 = 120 m<\/p>\n

7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Due to stoppages, the bus travelled 30 km less in an hour.
\nTime required to travel 30 km without stoppages = $= \\dfrac{30}{75}\\times60 = 24$ min<\/p>\n

8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let the speed of boat upstream = 5x km\/hr
\nLet the speed of boat downstream = 14x km\/hr<\/p>\n

Speed of boat in still water = (14x+5x)\/2= 19x\/2 km\/hr
\nSpeed of stream = Speed of boat in still water – Speed of boat upstream = (19x\/2) – 5x = 9x\/2 km\/hr
\nGiven Speed of stream = 18 km\/hr
\n9x\/2 = 18
\n\u21d2 9x = 36
\n\u21d2 x = 4
\nThen, Speed of boat in still water = 19*4\/2 = 38 km\/hr<\/p>\n

9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the onward and return distance be 60 km (LCM of 60 and 30)
\nTime taken for onward journey = 60\/60 = 1 hour
\nTime taken for return journey = 60\/30 = 2 hours
\nTotal time = 3 hours
\nTotal distance = 60+60 = 120 km
\nAverage speed = Total Distance\/Total Time = 120\/3 = 40 km\/hr<\/p>\n

\"RRB<\/a><\/p>\n

10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let the onward distance be d km
\nAs the distance is same, return distance = d km
\nTotal distance = 2d km
\nTime to travel from his home to his office $=\\dfrac{d}{40}$ hr
\nTime to travel from his office to his home $ =\\dfrac{d}{60}$ hr
\nAverage speed = $\\dfrac{Total distance}{Total time}<\/p>\n

$= \\dfrac{2d}{(\\dfrac{d}{40})+(\\dfrac{d}{60})$
\n$ = \\dfrac{2d}{(\\dfrac{5d}{120}) = 48 km\/hr<\/p>\n

11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Given that the distance $= 480$ km
\nLet the speed of the train $= x$ km\/hr
\nThen, time in which it can travel $240$ km $= \\dfrac{480}{x}$ hr
\nNow, Speed is reduced by $8$ km\/hr
\nReduced speed $= (x-8)$ km\/hr
\nTime in which it can travel $240$ km at reduced speed $= \\dfrac{480}{(x-10)}$
\nDifference in time = 3 hours<\/p>\n

\u21d2 $\\dfrac{480}{(x-8)} – \\dfrac{480}{x} = 3$<\/p>\n

\u21d2 $\\dfrac{480(x-x+8)}{x^2-8x} = 3$<\/p>\n

\u21d2 $3640 = 3(x^2-8x)$
\n\u21d2 $x^2-8x-1280 = 0$
\n\u21d2 $x^2-40x+32x-1280 = 0$
\n\u21d2 $x(x-40)+32(x-40) = 0$
\n\u21d2 $(x-40)(x+32) = 0$
\n$x-40 = 0$ | $ x+32 = 0$
\n$x = 40$ | $x = -32$
\nSpeeds cannot be negative.
\nHence, Initial speed $= 40$ km\/hr<\/p>\n

12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Given that the distance $= 240$ km
\nLet the speed of the train $= x$ km\/hr
\nThen, time in which it can travel $240$ km $= \\dfrac{240}{x}$ hr
\nNow, Speed is reduced by $10$ km\/hr
\nReduced speed $= (x-10)$ km\/hr
\nTime in which it can travel $240$ km at reduced speed $= \\dfrac{240}{(x-10)}$
\nDifference in time = 4 hours<\/p>\n

\u21d2 $\\dfrac{240}{(x-10)} – \\dfrac{240}{x} = 4$<\/p>\n

\u21d2 $\\dfrac{240(x-x+10)}{x^2-10x} = 4$<\/p>\n

\u21d2 $2400 = 4(x^2-10x)$
\n\u21d2 $x^2-10x-600 = 0$
\n\u21d2 $x^2-30x+20x-600 = 0$
\n\u21d2 $x(x-30)+20(x-30) = 0$
\n\u21d2 $(x+20)(x-30) = 0$
\nx+20 = 0 | x-30 = 0
\nx = -20 | x = 30
\nSpeeds cannot be negative.
\nHence, Initial speed = 30 km\/hr<\/p>\n

13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let the speed of boat upstream = x km\/hr
\nSpeed of boat in still water = 2x km\/hr
\nSpeed of stream = 2x-x = x km\/hr
\nSpeed of boat downstream = 2x+x = 3x km\/hr
\nRequired ratio = 3x:x = 3:1<\/p>\n

14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Given, Speed of boat downstream = 54 km\/hr
\nSpeed of current = 14 km\/hr
\nSpeed of boat in still water = 54-14 = 40 km\/hr
\nSpeed of boat upstream = Speed of boat in still water – Speed of current = 40-14 = 26 km\/hr<\/p>\n

15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let the speed of stream = x km\/hr
\nSpeed of boat in still water = 3x km\/hr
\nThen, Speed of boat upstream = 3x-x = 2x km\/hr
\nSpeed of boat downstream = 3x+x = 4x km\/hr
\nRequired ratio = 2x : 4x = 1:2<\/p>\n

16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the speed of boat upstream = 2x km\/hr
\nSpeed of boat downstream = 4x km\/hr
\nThen speed of boat in still water = (2x+4x)\/2 = 6x\/2 = 3x km\/hr
\nSpeed of current = 4x-3x = x km\/hr
\nRequired ratio = 3x:x = 3:1<\/p>\n

17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the length of the train be \u2018x\u2019 m
\nTotal distance it needs to travel to cross the platform = (180+x) m
\nEquating speeds
\n$\\dfrac{180+x}{16} = \\dfrac{x}{4}
\n\u21d2 x+180 = 4x
\n\u21d2 3x = 180
\n\u21d2 x = 60
\nHence, Length of the train = 60 m<\/p>\n

18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Total distance travelled by the train to cross the platform = 120+160 = 280 m
\nTime = 14 seconds
\nSpeed of the train in m\/sec = 280\/14 = 20 m\/sec
\nSpeed of the train in km\/hr = $\\dfrac{20\\times18}{5} = 72 km\/hr$<\/p>\n

19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let x be the length of the train,<\/p>\n

Speed = 90 kmph = 90*(5\/18) = 25 m\/s<\/p>\n

Distance = Speed*time<\/p>\n

x+600 = 25*(56) = 1400<\/p>\n

x = 1400 – 600 = 800m<\/p>\n

So the answer is option D.<\/p>\n

20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Speed = distance\/time = 600\/24 = 25 m\/s = 25*(18\/5) = 5*18 = 90 kmph<\/p>\n

So the answer is option D.<\/p>\n

21)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let the number of children = c<\/p>\n

Number of children is proportional to quantity of chocolates and inversely proportional to number of days.<\/p>\n

==> $\\frac{(12)(6)}{48} = \\frac{(c)(6)}{72}$<\/p>\n

==> $3\/2 = c\/12$<\/p>\n

==> $ c = 18$<\/p>\n

So the answer is option C.<\/p>\n

22)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let the speed of train X is x kmph<\/p>\n

Relative speed = 90-x kmph (if train Y is faster than train X)<\/p>\n

Speed = distance\/time = (850+950)\/(6*60) = 1800\/360 = 5 m\/s = 5*(18\/5) = 18 kmph<\/p>\n

90 – x = 18 kmph<\/p>\n

x = 90 – 18 = 72 kmph<\/p>\n

So the answer is option A.<\/p>\n

Case 2 :(If train Y is slower than train X)<\/p>\n

x – 90 = 18, then x = 90 + 18 = 108 kmph<\/p>\n

\"RRB<\/a><\/p>\n

23)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

6 hrs 30 mins = 6.5 hrs<\/p>\n

Speed = $\\frac{distance}{time} = \\frac{2210}{6.5} = 340 kmph$<\/p>\n

SO the answer is option D.<\/p>\n

24)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Time in case 1 = x\/60 hrs<\/p>\n

Time in case 2 = 2x\/40 = x\/20 hrs<\/p>\n

Total time = x\/60 + x\/20 = x\/15 hrs<\/p>\n

Total distance = x+2x = 3x km<\/p>\n

Average speed = total distance\/ total time = (3x)\/(x\/15) = 45 kmph<\/p>\n

So the answer is option A.<\/p>\n

25)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

50.4 kmph = 50.4*(5\/18) = 14 m\/s<\/p>\n

Distance = speed*time<\/p>\n

x+200 = (14)*(50)<\/p>\n

x+200 = 700<\/p>\n

x = 500m<\/p>\n

So the answer is option D.<\/p>\n

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We hope this Time & Speed questions for RRB NTPC Exam will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Time & Speed Questions for RRB NTPC PDF Download RRB NTPC Time & Speed Questions and Answers PDF. Top 25 RRB NTPC\u00a0 Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam Question 1:\u00a0P can do a work in 10 days, Q can do the same work in […]<\/p>\n","protected":false},"author":41,"featured_media":26105,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1603],"tags":[489,491,42,1619],"class_list":{"0":"post-26103","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-railways","10":"category-rrb-ntpc","11":"tag-railway-exam","12":"tag-rrb","13":"tag-rrb-mock-test","14":"tag-rrb-ntpc-2019"},"better_featured_image":{"id":26105,"alt_text":"Time & Speed Questions For RRB NTPC PDF","caption":"Time & Speed Questions For RRB NTPC PDF","description":"Time & Speed Questions For RRB 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